Question

Consider a random sample of 16 measurements obtained from a normally distributed population with =450 and s= 60. Construct a 90% confidence interval
OA (429.9, 470.1)
B. (430.8, 469.2)
Oc.(447.3, 452.6)
OD. (423.7, 476.3)

Answer 1

Solution:

Given that sample mean Xbar = 450, and sample standard deviation = s = 60, sample size = n = 16

Now 90 % confidence interval is given by, (Xbar t_0.90,df * s / sqrt(n))

Now for df = n - 1 = 16 - 1 = 15, value of   t_{0.90, 15} = 1.75, using student's t-distribution for confidence c = 0.9 and d.f. = 15.

Now plug the values in the formula we will get,

( 450 ± 1.75 * 60 / sqrt(16)) = (450 +1.75*15 , 450 - 1.75*15) = (423.7, 476.3)

Hence option D. (423.7, 476.3) is correct.

Best Luck !